A person, standing exactly midway between two towers, observes the top of the two towers at angle of elevation of 22.5° and 67.5°. What is the ratio of the height of the taller tower to the height of the shorter tower? (Given that tan 22.5° = 2 - 1 )

Answer: Option C

Let ED be the taller tower and AB be the shorter tower.
Let C be the point of observation
Given that ∠ ACB = 22.5° and ∠ DCE = 67.5°
Given that C is the midpoint of BD
Hence, BC = CD
$$\begin{array}{rl}& \text{From}\text{the}\text{right}\Delta ABC,\\ & \tan 22.5^{\circ }=\frac{AB}{BC}.....\left(eq:1\right)\\ & \text{From}\text{the}\text{right}\Delta CDE,\\ & \tan 67.5^{\circ }=\frac{ED}{CD}.....\left(eq:2\right)\\ & \frac{\left(eq:2\right)}{\left(eq:1\right)}\Rightarrow \frac{\tan {67.5}^{\circ }}{\tan {22.5}^{\circ }}=\frac{\left(\frac{ED}{CD}\right)}{\left(\frac{AB}{BC}\right)}\\ & =\frac{ED}{AB}\left(\because CD=BC\right)\\ & \Rightarrow \frac{\tan \left({90}^{\circ }-{22.5}^{\circ }\right)}{\tan {22.5}^{\circ }}=\frac{ED}{AB}\\ & \Rightarrow \frac{\cot {22.5}^{\circ }}{\tan {22.5}^{\circ }}=\frac{ED}{AB}\\ & \left[\because \tan \left(90-\theta \right)=\cot \theta \right]\\ & \Rightarrow \frac{\left(\frac{1}{\tan {22.5}^{\circ }}\right)}{\tan {22.5}^{\circ }}=\frac{ED}{AB}\\ & \left[\because \cot \theta =\frac{1}{\tan \theta }\right]\\ & \Rightarrow \frac{ED}{AB}=\frac{1}{{\left(\tan {22.5}^{\circ }\right)}^2}\\ & =\frac{1}{{\left(\sqrt{2}-1\right)}^2}\\ & ={\left(\frac{1}{\sqrt{2}-1}\right)}^2\\ & ={\left[\frac{\left(\sqrt{2}+1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}\right]}^2\\ & ={\left[\frac{\left(\sqrt{2}+1\right)}{\left(2-1\right)}\right]}^2\\ & ={\left[\frac{\left(\sqrt{2}+1\right)}{1}\right]}^2\\ & ={\left(\sqrt{2}+1\right)}^2\\ & =\left(2+2\sqrt{2}+1\right)\\ & =\left(3+2\sqrt{2}\right)\\ & \text{Required}\text{Ratio}\\ & =ED:AB\\ & =\left(3+2\sqrt{2}\right):1\end{array}$$